arXiv:1112.4080v1 [nlin.SI] 17 Dec 2011
Stability of periodic traveling waves for the
quadratic and cubic nonlinear Schr¨
odinger equations
Sevdzhan Hakkaev
1, Iliya D. Iliev
2and Kiril Kirchev
2December 20, 2011
1Faculty of Mathematics and Informatics,
Shumen University, 9712 Shumen, Bulgaria
2Institute of Mathematics and Informatics,
Bulgarian Academy of Sciences, 1113 Sofia, Bulgaria
Abstract
We study the existence and stability of periodic traveling-wave solutions for the quadratic and cubic nonlinear Schr¨odinger equations in one space dimen-sion.
1
Introduction
In this work we consider the nonlinear Schr¨odinger equation (NLS)
iut+ uxx+ |u|pu = 0. (1.1)
This equation appears in various problems, modeling many phenomena such as the behavior of a non-ideal Bose gas with a weak particle interaction, the spreading of the heat impulse in solids, the Langmuir waves in a plasma, etc. [17, 18].
Our principal aim is to study the the orbital stability of the family of periodic traveling-wave solutions
u = ϕ(x, t) = ei(vx+(ω−v2)t)r(x − 2vt). (1.2) where r(y) is a real-valued T -periodic function and v, ω ∈ R are parameters, for quadratic (p=1) and cubic (p=2) nonlinear Schr¨odinger equation.The problem of the stability of solitary waves for nonlinear dispersive equations goes back to the works of Benjamin [4] and Bona [5] (see also [1, 15, 16]). A general approach for investigating the stability of solitary waves for nonlinear equations having a group of symmetries was proposed in [8]. The existence and stability of solitary wave solutions for equation (1.1) has been studied in [19]. Recently in [3], the authors developed a
complete theory on the stability of cnoidal waves for the KdV equation. Other new explicit formulae for the periodic traveling waves based on the Jacobi elliptic func-tions, together with their stability, have been obtained in [2, 9, 10] for the nonlinear Schr¨odinger equation, modified KdV equation, complex modified KdV equation, and generalized BBM equation. In [11], the stability of periodic traveling wave solutions of BBM equation which wave profile stays close to the constant state u = (c − 1)1/p
is considered.
In this paper, we prove stability of the periodic traveling waves (1.2) not oscillating around zero (r 6= 0) for the quadratic and the cubic NLS. Our main results are formulated as Theorem 2.1 and Theorem 3.1 in Sections 2 and 3 below. We base our analysis on some appropriate invariant laws. Our approach is to verify that ϕ is a minimizer of a properly chosen functional M which is conservative with respect to time over the solutions of (1.1). We consider the L2-space of T -periodic functions in
x ∈ R, with a norm ||.|| and a scalar product h., .i. To establish that the orbit O = {eiηϕ(· − ξ, t) : (ξ, η) ∈ [0, T ] × [0, 2π]}
is stable, we take
u(x, t) = eiηϕ(x − ξ, t) + h(x, t) = eiζ[r(x − ξ − 2vt) + h1+ ih2]
and express the leading term of M(u) − M(ϕ) as hL1h1, h1i + hL2h2, h2i where Li are
second-order selfadjoint differential operators in L2[0, T ] with potentials depending
on r and satisfying L1r′ = L2r = 0. The proof of orbital stability requires that zero
is the second eigenvalue of L1 and the first one of L2.
Recall that the quadratic and cubic nonlinear Schr¨odinger equations are globally well-posed in Hs(T), for s ≥ 0 [6].
The paper is organized as follows. We consider the quadratic and the cubic cases in Sections 2 and 3 respectively. In Appendices 1 and 2, some facts from the theory of complete Abelian integrals (e.g. Picard-Fuchs equations, polynomial moduli) are used in order to derive several inequalities we needed during the proof our main results.
2
Existence and stability of periodic traveling waves
for the quadratic Schr¨
odinger equation
Consider the equation
iut+ uxx+ |u|u = 0, (2.1)
where u is a complex-valued function.
We are looking for a solution of equation (2.1) in the form (1.2) where r is real-valued. For r one obtains the equation
r′′− ωr + r|r| = 0. (2.2) Therefore, r′2− ωr2+ 2 3r 2 |r| = c (2.3)
and r is periodic provided that the level set H(x, y) = c of the Hamiltonian system dH = 0, with
H(x, y) = y2− ωx2+ 2
3x
2|x|,
contains a periodic trajectory (an oval). The level set H(x, y) = c contains two periodic trajectories if ω > 0, c ∈ (−13ω3, 0) and a unique periodic trajectory if
ω ∈ R, c > 0. Under these conditions, equation (2.3) becomes H(r, r′) = c and its
solution r is periodic of period T = T (ω, c).
Below, we consider the case c < 0. Then either r < 0 (the left case) or r > 0 (the right case). To express r through elliptic functions, we denote by r0 > r1 > 0 the
positive solutions of 23ρ3− ωρ2− c = 0. Then r
1 ≤ |r| ≤ r0 and one can rewrite (2.3)
as
r′2 = 2
3(|r| − r1)(r0− |r|)(|r| + r0+ r1 − 3
2ω). (2.4)
Therefore 2r0 + r1 > r0 + 2r1 > 32ω. Introducing a new variable s ∈ (0, 1) via
|r| = r1+ (r0− r1)s2, we transform (2.4) into
s′2= α2(1 − s2)(k′2+ k2s2) where α, k, k′ are positive constants (k2+ k′2= 1) given by
α2 = 4r0+ 2r1− 3ω 12 , k 2 = 2r0− 2r1 4r0+ 2r1− 3ω , k′2 = 2r0+ 4r1− 3ω 4r0+ 2r1− 3ω . Therefore |r(x)| = r1+ (r0− r1)cn2(αx; k). (2.5) T = 2K(k) α = 4√4 1 − k2+ k4K(k) √ ω , k ∈ (0, 1), T ∈ I = 2π √ ω, ∞ . (2.6)
As usual, here and below, K(k) and E(k) denote the complete elliptic integrals of the first and the second kind in a Legendre form. Let us recall for later use the system they satisfy:
kK′ = E
1 − k2 − K, kE′ = E − K.
Lemma 2.1. For any ω > 0 and T ∈ I, there is a constant c = c(ω) such that the periodic traveling-wave solution (2.5) determined by H(r, r′) = c(ω) has a period T .
The function c(ω) is differentiable.
Proof. The statement follows from the implicit function theorem. It is easily seen that the period T is a strictly increasing function of k:
d dk( 4 √ 1 − k2+ k4K(k)) = k(2k2−1)K(k)+2(1−k2+k4)K′(k) 2(1−k2 +k4 )3/4 = 2(1−k2+k4)E(k)+(1−k2)(k2−2)K(k) 2k(1−k2 )(1−k2 +k4 )3/4 > 0.
Given ω and c in their range, consider the functions r0(ω, c), r1(ω, c), k(ω, c) and
T (ω, c) given by the formulas we derived above. We obtain ∂T ∂c = dT dk dk dc = 1 2k dT dk d(k2) dc . Further, using that k2 = 3 2r0−ω
4r0+2r1−3ω − 1 and 2 3r 3 0− ωr20 = 23r 3 1− ωr21, we have d(k2) dc = 3 (4r1− 2ω)∂r∂c0 − (4r0− 2ω)∂r∂c1 (4r0+ 2r1− 3ω)2 = 3ω 2(r 1− r0) 4(r2 0 − ω2)(r21− ω2)(4r0+ 2r1− 3ω)2 .
We see that ∂T (ω, c)/∂c 6= 0, therefore the implicit function theorem yields the result. ✷
Equation (2.1) has the following conservation laws
Q(u) = i Z T 0 uxudx, P (u) = Z T 0 |u| 2dx, E(u) = Z T 0 (|u x|2− 2|u| 3 3 )dx. Let us consider the functional
M(u) = E(u) + (ω + v2)P (u) − 2vQ(u). Next we introduce the pseudometric
d(u, ϕ) = inf
(η,ξ)∈[0,2π]×[0,T ]||u(x, t) − e iη
ϕ(x − ξ, t)||1. (2.7)
For a fixed q > 0, we denote d2q(u, ϕ) = inf
(η,ξ)∈[0,2π]×[0,T ](||ux(x, t) − e iηϕ
x(x − ξ, t)||2+ q||u(x, t) − eiηϕ(x − ξ, t)||2).
(2.8) Clearly, the infimum in (2.8) is attained at some point (η, ξ) in the square [0, T ]×[0, T ]. Moreover, for q ∈ [q1, q2] ⊂ (0, ∞), (2.8) is a pseudometric equivalent to (2.7).
Lemma 2.2. The metric dq(u, ϕ) is a continuous function of t ∈ [0, ∞).
Proof. The proof of the lemma is similar to the proof of Lemmas 1, 2 in [5] ✷. Now, we can formulate our main result in this section.
Theorem 2.1. Let ϕ be given by (1.2), with r 6= 0. For each ε > 0 there exists δ > 0 such that if u(x, t) is a solution of (2.1) and d(u, ϕ)|t=0 < δ, then d(u, ϕ) < ε ∀t ∈ [0, ∞).
The crucial step in the proof will be to verify the following statement.
Proposition 2.1. There exist positive constants m, q, δ0 such that if u is a periodic
solution of (2.1), u(x, t) = u(x + T, t), P (u) = P (ϕ) and dq(u, ϕ) < δ0, then
Proof. In order to estimate ∆M = M(u) − M(ϕ), we set u(x, t) = eiηϕ(x − ξ, t) + h(x, t), ζ = v(x − ξ) + (ω − v2)t + η, F (s) = −2 3|e iη ϕ + hs|3. We have 2 3(−|u| 3 + |ϕ|3) = F (1) − F (0) = F′(0) + F′′(0) 2 + F′′(s) − F′′(0) 2 where
0 ≤ s ≤ 1, F′(0) = −2|ϕ|Re(eiηϕh), F′′(0) = −|ϕ|Re(h2e−2iζ) + 3|h|2. Integrating by parts in the terms containing hx and hx, we obtain
∆M = M(u) − M(ϕ)
= 2ReR0T eiη[−ϕ
xx+ (ω + v2− |ϕ|)ϕ + 2ivϕx]hdx
+R0T [|hx|2 + (ω + v2− 23|ϕ|)|h|2−|ϕ|2 Re(e−2iζh2) − 2ivhhx]dx
−R0T
F′′(s)−F′′(0)
2 dx
= I1+ I2+ I3.
Using that r(x) satisfies the equation (2.2) we obtain that I1 = 0.
Let h(x, t) = (h1+ ih2)eiζ, where h1 and h2 are real periodic functions with period
T . Then we have |h|2= h2 1+ h22 |hx|2 = h21x+ h22x+ 2v(h1h2x− h1xh2) + v2(h21 + h22) Re(e−2iζh2) = h2 1 − h22 RT 0 hhxdx = i RT 0 (h1xh2− h1h2x− vh21− vh22)dx. (2.10)
Finally, for I2 we obtain
I2 = Z T 0 [h21x + (ω − 2|r|)h21]dx + Z T 0 [h22x + (ω − |r|)h22]dx = M1 + M2
Consider in [0, T ] = [0, 2K(k)/α] the formal differential operators L1 = − d2 dx2 + (ω − 2|r|), L2 = − d2 dx2 + (ω − |r|),
supplied with periodic boundary conditions. By the above formulas, r0− r1 = 6α2k2,
2r0− ω = 4α2(1 + k2). Taking y = αx as an independent variable in L1, one obtains
L1 = α2Λ1 with an operator Λ1 in [0, 2K(k)] given by
Λ1 = − d2 dy2 + α −2[ω − 2(r 1+ (r0− r1)cn2(y; k))] = − d 2 dy2 + ω − 2r0 α2 + 2(r0− r1) α2 sn 2(y; k) = − d 2 dy2 − 4(1 + k 2) + 12k2sn2(y; k).
The spectral properties of the operator Λ1 in [0, 2K(k)] are well known. The first
three(simple) eigenvalues and corresponding eigenfunctions of Λ1 are
µ0 = k2− 2 − 2 √ 1 − k2+ 4k4 < 0, ψ0(y) = dn(y; k)[1 − (1 + 2k2− √ 1 − k2+ 4k4)sn2(y; k)] > 0 µ1 = 0
ψ1(y) = dn(y; k)sn(y; k)cn(y; k) = 12dydcn2(y; k)
µ2 = k2− 2 + 2 √ 1 − k2+ 4k4 > 0 ψ2(y) = dn(y; k)[1 − (1 + 2k2+ √ 1 − k2+ 4k4)sn2(y; k)].
Since the eigenvalues of L1 and Λ1 are related by λn = α2µn, it follows that the first
three eigenvalues of the operator L1, equipped with periodic boundary condition on
[0, 2K(k)] are simple and λ0 < 0, λ1 = 0, λ2 > 0. The corresponding eigenfunctions
are ψ0(αx), ψ1(αx) = const.r′ and ψ2(αx).
In a similar way, with L2 = α2Λ2, one obtains in [0, 2K(k)]
Λ2= −
d2
dy2 − 2(1 + k
2) + 6k2sn2(y; k) + ω/2α2.
To express ω through α and k, one should take into account the fact that in the cubic equation we used to determine r0 and r1, the coefficient at ρ is zero. Therefore,
r0r1+ (r0 + r1)(32ω − r0− r1) = 0.
As r0 = 2α2+ 2α2k2+12ω, r1 = 2α2− 4α2k2+12ω, after replacing these values in the
above equation one obtains ω2 = 16α4(1 − k2 + k4). Since ω > 0, we finally obtain
Λ2 = −
d2
dy2 + 2(−1 − k 2+√
On the other hand, (2.5) yields
|r| = 2α2[1 + k2+√1 − k2+ k4− 3k2sn2(y; k)].
The first three eigenvalues and corresponding eigenfunctions of Λ2 are as follows:
ǫ0 = 0, υ0 = r, ǫ1 = 2 − k2+ 2 √ 1 − k2+ k4, υ 1 = dn′(y; k) ǫ2 = 4 √ 1 − k2+ k4, υ 2 = 1 + k2− √ 1 − k2+ k4− 3k2sn2(y; k). Estimates for M2.
From the above explanations we know that when considered in [0, T ], the operator L2
has an eigenfunction r corresponding to zero eigenvalue and the rest of the spectrum is contained in (α2λ
1, ∞).
In the formulas which follow, we take r = r(¯x) with an argument ¯x = x − ξ − 2vt. The values of ξ and η are chosen so that the infimum in (2.8) is attained at that point. Therefore the derivative of d2q(u, ϕ) with respect to η is equal to zero. Together with
(2.2), this yields
0 = iR0T hxe−iηϕx− ϕxeiηhx+ q(he−iηϕ − ϕeiηh)dx
= 2ImR0T (−ϕxx+ qϕ)eiηhdx
=R0T [(v2− ω + q + |r|) rh 2+ 2vr′h1]dx. (2.11) We set h2 = βr(¯x) + θ, RT 0 θrdx = 0. Substituting in (2.11), we obtain β||r||2 v2− ω + q + ||r 3/2||2 ||r||2 + Z T 0 [θr|r| + 2vr ′h 1]dx = 0.
Using that ||r||r||3/22||2 > ω (see estimate A1 of Appendix 1), we obtain the estimate
|β| ||r|| ≤ R0T(θr|r| + 2vr′h1)dx (v2+ q)||r|| ≤ ||r 2|| · ||θ|| + 2|v| ||r′|| · ||h 1|| (v2+ q)||r|| ≤ m0(||θ|| + ||h1||),
where m0 = 2m1(v, ω)/(q + ω2) and m1(v, ω) = max c∈[−1 3ω 3,0] ||r2|| ||r||, 2|v| ||r′|| ||r|| , 2|| |r|r′|| ||r′|| , |v||| |r|r − ωr|| ||r′|| .
Clearly, the first and the third terms are uniformly bounded for ω fixed. The bound-edness of the second and the fourth ones follows from the estimates in D1 of Appendix 1.(The third and fourth terms are included for later use.) We will use below that for v and ω fixed, m0 → 0 when q → ∞. Further,
||h2|| ≤ |β| ||r|| + ||θ|| ≤ m0(||θ|| + ||h1||) + ||θ|| = (m0+ 1)||θ|| + m0||h1||. Hence, we obtain ||θ||2 ≥ ||h2|| 2 2(m0+ 1)2 − m0 m0+ 1 2 ||h1||2. (2.12)
Since L2r = 0 and hθ, ri = 0, then from the spectral properties of the operator L2, it
follows
M2 = hL2h2, h2i = hL2θ, θi ≥ α2ǫ1||θ||2.
From here and (2.12), one obtains
M2 ≥ α2ǫ 1 2(m0+ 1)2||h 2||2− α2ǫ 1m20 (m0+ 1)2||h 1||2. (2.13) Estimates for M1. We set h1 = γ1ψ0(¯x) + γ2r′(¯x) + θ1, r(¯x) = νψ0(¯x) + ψ, (2.14) where hθ1, ψ0i = hθ1, r′i = hψ, ψ0i = hψ0, r′i = hψ, r′i = 0 (2.15)
and γ1, γ2 and ν are some constants. By (2.15), we have
M1(h1) = hL1h1, h1i = γ12λ0hψ0, ψ0i + hL1θ1, θ1i.
Therefore, from spectral properties of the operator L1 it follows
M1(h1) ≥ γ12λ0||ψ0||2+ λ2||θ1||2. (2.16)
The fundamental difficulty in the estimate of M1 is the appearance of the negative
term γ2
1λ0||ψ0||2. Below, we are going to estimate it. From the condition
P (u) = Z T 0 |h + e iωη ϕ(x − ξ, t)|2dx = P (ϕ) we obtain ||h||2 = 2Re Z T 0 eiωηϕ(x − ξ, t)hdx = −2 Z T 0 rh1dx.
Then using (2.14), we have −12||h||2 = νγ1||ψ0||2+ Z T 0 ψθ1dx and therefore γ12||ψ0||2 = 1 ν2||ψ 0||2 1 2||h|| 2+ Z T 0 ψθ1dx 2 . (2.17) From (2.17), we obtain γ12||ψ0||2 ≤ 1 ν2||ψ 0||2 1 + d 4 ||h|| 4+d + 1 d ||ψ|| 2 ||θ1||2 , (2.18)
where d is a positive constant which will be fixed later. Using (2.17) and (2.16), we derive the inequality
M1 ≥ λ2+ λ0(1 + 1 d) ||ψ||2 ν2||ψ 0||2 ||θ1||2+ λ0(1 + d) 4ν2||ψ 0||2||h|| 4 (2.19)
Below, we will denote by Cm, Dm positive constants. By using (2.18) and (2.19),
we derive the inequality M1 ≥ λ2+ λ0(1 + 1d) ||ψ|| 2 ν2 ||ψ0||2 ||θ1||2+ λ 0(1+d) 4ν2 ||ψ0||2||h|| 4 ≥ C1λ2||θ1||2− D1||h||4 (2.20)
(see the estimates in point C1 of the Appendix 1).
We denote ϑ = h1− γ2r′(¯x) = γ1ψ0(¯x) + θ1. Then from (2.14), (2.20), we have
||ϑ||2 = γ2 1||ψ0||2+ ||θ1||2 ≤ 1 + (d+1)||ψ||dν2 2 ||ψ0||2 ||θ1||2+ 4ν21+d ||ψ0||2||h|| 4 ≤ C2||θ1||2+ D2||h||4. Then ||θ1||2 ≥ ||ϑ|| 2 C2 − D2||h||4 C2|a| 1 2 and hence, by (2.20), M1 ≥ C1λ2 C2 ||ϑ|| 2 − C1λ2D2C+ C2D1 2 ||h|| 4. (2.21)
After differentiating (2.8) with respect to ξ and using (2.2), we obtain 0 = 2ReR0T eiωη(ϕ xxhx+ qϕxh)dx = 2ReR0T [(r′′+ 2ivr′− v2r)(h 1x− ih2x− ivh1− vh2) + q(r′+ ivr)(h1− ih2)]dx = 2R0T [(−ω + 2|r| + 3v2+ q)r′h 1+ v(v2− 3ω + 3|r| + q)rh2]dx.
From (2.11), we have Z T 0 qrh2dx = − Z T 0 [(v2− ω + |r|)rh2+ 2vr′h1]dx
and replacing in the above equality, we obtain Z T
0 [(−ω + 2|r| + v
2+ q)r′h
1+ v(2|r| − 2ω)rh2]dx = 0.
Substituting h1 = γ2r′(¯x) + ϑ in the above equality and using the orthogonality
condition hr′, ϑi = hr′, γ 1ψ0+ θ1i = 0, we obtain γ2||r′||2 −ω + v2+ q + 2|| √ rr′||2 ||r′||2 + Z T 0 [v(2|r| − 2ω)rh 2+ 2|r|r′ϑ]dx = 0.
As 2||||r√rr′||2′||2 ≥ ω (see estimate B1 from Appendix 1),
|γ2| ||r′|| ≤ R0T [v(2|r| − 2ω)rh2+ 2|r|r′ϑ]dx (v2+ q)||r′|| ≤ 2|v| |||r|r − ωr|| · ||h(v2 2|| + 2|||r|r′|| · ||ϑ|| + q)||r′|| ≤ m0(||ϑ|| + ||h2||). Hence ||h1|| ≤ |γ2| ||r′|| + ||ϑ|| ≤ (m0+ 1)||ϑ|| + m0||h2||, which yields ||ϑ||2≥ ||h1|| 2 2(m0+ 1)2 − m0 m0+ 1 2 ||h2||2.
Replacing in (2.21), we finally obtain
M1 ≥ C1λ2 C22(m0+ 1)2||h1|| 2 − C1λ2m 2 0 C2(m0+ 1)2||h2|| 2 −C1λ2DC2+ C2D1 2 |h|| 4. (2.22)
The estimate for ∆M.
From (2.13) and (2.22), we have (fixing q large and therefore m0 small enough)
M1+ M2 ≥ C1λ2−2C2α2ǫ1m20 2C2(m0+1)2 ||h1||2+ C2α2ǫ1−2C1λ2m20 2C2(m0+1)2 ||h2||2 −C1λ2D2+C2D1 C2 ||h|| 4 ≥ C 3||h||2− D3||h||4.
On the other hand, estimating directly I2 from below (for this purpose we use its
initial formula), we obtain I2 ≥ ||hx||2+ RT 0 (ω + v 2− 2|ϕ|)|h|2dx − 2|v|RT 0 |h| · |hx|dx ≥ ||hx||2+ (ω + v2− 2 max |r|)||h||2− 2v2||h||2− 12||hx||2 = 1 2||hx|| 2+ (ω − v2− 2r 0)||h||2. Let 0 < m < 1 2. We have ∆M = 2mI2+ (1 − 2m)(M1+ M2) + I3 ≥ m||hx||2+ 2m(ω − v2− 2r0)||h||2+ (1 − 2m)(C3||h||2− D3||h||4) + I3 ≥ m||hx||2+ [(2m(ω − v2− 2r0) + (1 − 2m)C3]||h||2− (1 − 2m)D3||h||4− |I3|.
We choose m, so that 2mq = (1 − 2m)C3+ 2m(ω − v2− 2r0), i.e.
2m = C3
q + C3− ω + v2+ 2r0
.
From the continuity of |z| and |z|e−2iargz, we have
|I3| <
mq 2 ||h||
2
From the inequality
|h|2 ≤ T1 Z T 0 |h| 2dx + 2 Z T 0 |h| 2dx Z T 0 |h x|2dx 1 2 we obtain |h|2 ≤ 1 T Z T 0 |h| 2dx +√qZ T 0 |h| 2dx + 1 √q Z T 0 |h x|2dx.
Hence for sufficiently large q, we obtain
max |h(x, t)|2 ≤ √2qd2q(u, ϕ)
and moreover ||h||2 ≤ q−1d2
q(u, ϕ). Consequently we can choose δ0 > 0, such that for
dq(u, ϕ) < δ0, we will have [max(4|a|
1
2|h| + |h|2) + (1 − 2m)D
3|a|
1
2]||h||2 ≤ qm. Finally, we obtain that if dq(u, ϕ) < δ0, then ∆M ≥ md2q(u, ϕ). Proposition 2.1 is
completely proved. ✷
Proof of Theorem 2.1. We split the proof of our main result into two steps. We begin with the special case P (u) = P (ϕ). Assume that m, q, δ0 have been selected
according to Proposition 2.1. Since ∆M does not depend on t, t ∈ [0, ∞), there exists a constant l such that ∆M ≤ ld2(u, ϕ)|
t=0. Below, we shall assume without loss of
generality that l ≥ 1, q ≥ 1. Let ε > 0, δ = min m lq δ0 2, m l 1/2 ε
and d(u, ϕ)|t=0 < δ. Then
dq(u, ϕ) ≤ q1/2d(u, ϕ)|t=0 <
δ0
2
and Lemma 2.2 yields that there exists a t0 > 0 such that dq(u, ϕ) < δ0 if t ∈ [0, t0).
Then, by virtue of Proposition 2.1 we have
∆M ≥ md2q(u, ϕ), t ∈ [0, t0).
Let tmax be the largest value such that
∆M ≥ md2q(u, ϕ), t ∈ [0, tmax).
We assume that tmax < ∞. Then, for t ∈ [0, tmax] we have
d2q(u, ϕ) ≤ ∆M m ≤ l md 2 (u, ϕ)|t=0< l mδ 2 ≤ δ 2 0 4.
Applying once again Lemma 2.2, we obtain that there exists t1 > tmax such that
dq(u, ϕ) < δ0, t ∈ [0, t1).
By virtue of the proposition, this contradicts the assumption tmax < ∞.
Conse-quently, tmax = ∞, ∆M ≥ md2q(u, ϕ) ≥ md2(u, ϕ), t ∈ [0, ∞). Therefore, d2(u, ϕ) ≤ ∆M m ≤ l mδ 2 < ε2 , t ∈ [0, ∞), which proves the theorem in the special case.
Now we proceed to release the restriction P (u) = ||u||2 = ||ϕ||2 = P (ϕ). We have
||ϕ|| = (16α3√1 − k2+ k4h(k2− 2 +√1 − k2+ k4)K(k) + 3E(k)i)1/2.
Below, we are going to apply a perturbation argument, freezing for a while the period T and the parameters ω, c in (2.3). We claim there are respective parameter values ω∗, c∗, and corresponding ϕ∗, r∗, α∗, k∗, see (2.2), (2.3) and (2.5), such that ϕ∗ has a
period T in x and moreover, ||ϕ∗|| = ||u||. By (2.9), we obtain the equations
2K(k∗)
α∗ − T = 0,
||r∗||2− ||u||2 = 0.
Moreover, one has ||ϕ∗|| = ||u|| and we could use the restricted result we established
above. As k = k∗(T, ||ϕ||), α = α∗(T, ||ϕ||), it remains to apply the implicit function
theorem to (2.23). Since the corresponding Jacobian determinant reads ∂ ∂k∗ 2K(k∗) α∗ ∂ ∂α∗ 2K(k∗) α∗ ∂ ∂k∗||r∗|| 2 ∂ ∂α∗||r∗|| 2 > 0 the needed properties are established.
0.2 0.4 0.6 0.8 1 10000 20000 30000 40000 50000 60000
Figure 1: Graph of the Jacobian determinant By (2.23) and our assumption, we have
K(k) α = K(k∗) α∗ = T 2. (2.24)
Next, choosing η = 2(ω∗− ω)t, ξ = 0, we obtain inequality
d2(ϕ∗, ϕ) ≤ (1 + v2)||r∗− r||2+ ||r∗′− r′||2.
Denote for while Φ(ρ) = 2α2(1 − 2k2+√1 − k2+ k4+ 8k2cn2(αx; k)), where k = k(ρ)
is determined from K(k) = 1
2ρT . Then r∗− r = Φ(α∗) − Φ(α) = (α∗− α)Φ′(ρ) with
some appropriate ρ. Moreover, |Φ′(ρ)| ≤ C
0 with constant independent of the values
with ∗ accent. Hence |r∗− r| ≤ C
0|α∗− α|. Similarly |r′∗− r′| ≤ C1|α∗− α|. All this,
together with (2.24) yields
d(ϕ∗, ϕ) ≤ C|α∗− α|C = 2C
T |K(k
∗) − K(k)| = 2C
T |K
′(k)||k∗− k|. (2.25)
Let ε > 0. From the inequalities
|||ϕ∗|| − ||ϕ||| = |||u|| − ||ϕ||| ≤ d(u, ϕ)|t=0< δ
it follows that
and, consequently, | ||r∗||2− ||r||2| < ||r||2δ
1, where δ1 = (1 + ||r||δ)2− 1.
On the other hand, we have (using (2.24) again) | ||r∗||2− ||r||2| = T273/2/2 (K(k∗))3/2p1 − k∗2+ k∗4h(k∗2− 2 +p1 − k∗2+ k∗4K(k∗) + 3E(k∗)i −(K(k))3/2√1 − k2+ k4(k2− 2 +√1 − k2+ k4K(k) + 3E(k) ≥ C2||k∗− k|, (2.26) Thus combining (2.25) and (2.26), we get
d(u, ϕ∗)|t=0≤ d(u, ϕ)|t=0+ d(ϕ, ϕ∗)|t=0 < δ + ||r||2Cδe 1 = δ0.
We select δ sufficiently small and apply the part of the theorem which has been already proved,
d(u, ϕ∗)|t=0< δ0 ⇒ d(u, ϕ∗) <
ε
2, t ∈ [0, ∞). Choosing an appropriate δ > 0, we obtain that
d(u, ϕ) ≤ d(u, ϕ∗) + d(ϕ, ϕ∗) < ε, for all t ∈ [0, ∞). Theorem 2.1 is completely proved. ✷
3
Existence and stability of periodic traveling waves
for the cubic Schr¨
odinger equation
Consider the cubic nonlinear Schr¨odinger equation
iut+ uxx+ |u|2u = 0, (3.1)
where u = u(x, t) is a complex-valued function of (x, t) ∈ R2.
Equation (3.1) possesses the following family of traveling-wave solutions
ϕ(x, t) = ei(vx+(ω−v2)t)r(x − 2vt), (3.2) where ω and v are real parameters and the real-valued function r(x) satisfies equation r′′− ωr + r3 = 0. (3.3) Integrating once again, we obtain
r′2− ωr2+ 1 2r
and r is a periodic function provided that the energy level set H(x, y) = c of the Hamiltonian system dH = 0,
H(x, y) = y2− ωx2+ 1 2x
4,
contains an oval (a simple closed real curve free of critical points). The level set H(x, y) = c contains two periodic trajectories if ω > 0, c ∈ (−1
2ω
2, 0) and a unique
periodic trajectory if ω ∈ R, c > 0. Under these conditions, the solution of (3.3) is determined by H(r, r′) = c and r is periodic of period T = T (ω, c).
Below, we are going to consider the case c < 0. Let us denote by r0 > r1 > 0
the positive roots of 12r4 − ωr2 − c = 0. Then, up to a translation, we obtain the
respective explicit formulas
r(z) = ∓r0dn(αz; k), k2 = r2 0− r12 r2 0 = −2ω + 2r 2 0 r2 0 , α = √r0 2, T = 2K(k) α . (3.5) Recall that K(k) and E(k) are, as usual, the complete elliptic integrals of the first and second kind in a Legendre form. By (3.5), one also obtains ω = (2 − k2)α2 and,
finally, T = 2 √ 2 − k2K(k) √ ω , k ∈ (0, 1), T ∈ I = 2π √ ω, ∞ . (3.6)
We take ω > 0, an energy level c ∈ (−12ω
2, 0) and let T be the (minimal) period
of r(x). Assume that v 6= 0 is chosen to satisfy vT/2π ∈ Z. Then there are two periodic solutions r of (3.3): left (r < 0) and right (r > 0) and the corresponding traveling-wave solution ϕ(x, t) is periodic with respect to x of period T .
Lemma 3.1. For any ω > 0 and T ∈ I, there is a constant c = c(ω) such that the periodic traveling-wave solution (3.5) determined by H(r, r′) = c(ω) has a period T .
The function c(ω) is differentiable. Proof. See [10], Lemma 3.1.
Equation (3.1) has the following conservation laws
Q(u) = i Z T 0 uxudx, P (u) = Z T 0 |u| 2dx, E(u) =˜ Z T 0 (|u x|2− |u| 4 2 )dx. Let us consider the functional
˜
M (u) = ˜E(u) + (ω + v2)P (u) − 2vQ(u).
Theorem 3.1. Let ϕ be given by (3.2), with r 6= 0. For each ε > 0 there exists δ > 0 such that if u(x, t) is a solution of (3.1) and d(u, ϕ)|t=0 < δ, then d(u, ϕ) < ε ∀t ∈ [0, ∞).
Proposition 3.1. There exist positive constants m, q, δ0 such that if u is a periodic
solution of (3.1), u(x, t) = u(x + T, t), P (u) = P (ϕ) and dq(u, ϕ) < δ0, then
˜
M (u) − ˜M (ϕ) ≥ md2q(u, ϕ). (3.7)
Proof. In order to estimate ∆ ˜M = ˜M(u) − ˜M(ϕ), we set u(x, t) = eiηϕ(x − ξ, t) + h(x, t), ζ = v(x − ξ) + (ω + v2)t + η,
and integrating by parts in the terms containing hx and hx, we obtain
∆ ˜M = ˜M (u) − ˜M (ϕ)
= 2ReR0T eiη[−ϕ
xx + (ω + v2− |ϕ|2)ϕ + 2ivϕx]hdx
+R0T [|hx|2+ (ω + v2− 2|ϕ|2)|h|2− |ϕ|2Re(e−2iζh2) − 2ivhhx]dx
−1 2 RT 0 |h| 2(4Re(eiηϕh) + |h|2)dx = I1+ I2+ I3.
Using that r(x) satisfies the equation (3.3) we obtain that I1 = 0.
Let
h(x, t) = (h1 + ih2)eiζ,
where h1 and h2 are real periodic functions with period T . Using (2.10), for I2 we
obtain I2 = Z T 0 [h21x + (ω − 3r2)h21]dx + Z T 0 [h22x+ (ω − r2)h22]dx = ˜M1 + ˜M2
Introduce in L2[0, T ] the self-adjoint operators L
1 and L2 generated by the differential
expressions
L1 = −∂x2 + (ω − 3r2),
L2 = −∂x2 + (ω − r2),
(3.8)
with periodic boundary conditions in [0, T ].
We use now (3.5) and (3.6) to rewrite operators L1, L2 in more appropriate form.
From the expression for r(x) from (3.5) and the relations between elliptic functions sn(x), cn(x) and dn(x), we obtain
where y = αx.
It is well-known that the first five eigenvalues of Λ1 = −∂y2 + 6k2sn2(y, k), with
periodic boundary conditions on [0, 4K(k)], where K(k) is the complete elliptic inte-gral of the first kind, are simple. These eigenvalues and corresponding eigenfunctions are: ν0 = 2 + 2k2− 2 √ 1 − k2+ k4, φ 0(y) = 1 − (1 + k2− √ 1 − k2+ k4)sn2(y, k),
ν1 = 1 + k2, φ1(y) = cn(y, k)dn(y, k) = sn′(y, k),
ν2 = 1 + 4k2, φ2(y) = sn(y, k)dn(y, k) = −cn′(y, k),
ν3 = 4 + k2, φ3(y) = sn(y, k)cn(y, k) = −k−2dn′(y, k),
ν4 = 2 + 2k2+ 2 √ 1 − k2+ k4, φ 4(y) = 1 − (1 + k2+ √ 1 − k2+ k4)sn2(y, k).
It follows that the first three eigenvalues of the operator L1, equipped with periodic
boundary condition on [0, 2K(k)] (that is, in the case of left and right family), are simple and λ0 = α2(ν0− ν3) < 0, λ1 = α2(ν3− ν3) = 0, λ2 = α2(ν4− ν3) > 0. The
corresponding eigenfunctions are ψ0 = φ0(αx), ψ1 = r′(x), ψ2 = φ4(αx).
Similarly, for the operator L2 we have
L2 = α2[−∂2y + 2k2sn2(y, k) − k2]
in the case of left and right family. The spectrum of Λ2 = −∂y2+2k2sn2(y, k) is formed
by bands [k2, 1] ∪ [1 + k2, +∞). The first three eigenvalues and the corresponding
eigenfunctions with periodic boundary conditions on [0, 4K(k)] are simple and ǫ0 = k2, θ0(y) = dn(y, k),
ǫ1 = 1, θ1(y) = cn(y, k),
ǫ2 = 1 + k2, θ2(y) = sn(y, k).
From (3.4) it follows that zero is an eigenvalue of L2 and it is the first eigenvalue
in the case of left and right family, with corresponding eigenfunction r(x). Estimates for ˜M2.
As in Section 2, we use below r = r(¯x) with an argument ¯x = x − ξ − 2vt. From the explanations above, we know that when considered in [0, T ], the operator L2 has
an eigenfunction r corresponding to zero eigenvalue and the rest of the spectrum is contained in (α2, ∞).
The derivative of d2
q(u, ϕ) with respect to η at the point where the minimum is
attained is equal to zero. Together with (3.3), this yields Z T 0 v2− ω + q + r2rh2+ 2vr′h1 dx = 0 (3.9)
We set h2 = βr(¯x) + θ, RT 0 θrdx = 0. Substituting in (3.9), we obtain β||r||2 v2− ω + q + ||r 2||2 ||r||2 + Z T 0 [θr3+ 2vr′h1]dx = 0.
Using that 2||r||r||2||22 ≥ w (see estimate A2 of the Appendix2), we obtain the estimate
|β| ||r|| ≤ R0T(θr 3 + 2vr′h 1)dx (q + v2)||r|| ≤ ||r 3|| · ||θ|| + 2|v| ||r′|| · ||h 1|| (q + v2)||r|| ≤ m0(||θ|| + ||h1||), where m0 = 2m1(v, ω)/(q + ω2) and m1(v, ω) = max c∈[−1 2ω 2 ,0] ||r3|| ||r||, 2|v| ||r′|| ||r|| , 2|v| ||r3− ωr|| ||r′|| , 3||r2r′|| ||r′||
(the third and fourth item are included for later use). It is obvious that the first and the last fractions are bounded. For the second and the third ones, see estimates D2 in Appendix 2. We will use below that for v and ω fixed, m0 → 0 when q → ∞.
Further, ||h2|| ≤ |β| ||r|| + ||θ|| ≤ m0(||θ|| + ||h1||) + ||θ|| = (m0+ 1)||θ|| + m0||h1||. Hence, we obtain ||θ||2 ≥ ||h2|| 2 2(m0+ 1)2 − m0 m0+ 1 2 ||h1||2. (3.10)
Since L2r = 0 and hθ, ri = 0, then from the spectral properties of the operator L2, it
follows
M2 = hL2h2, h2i = hL2θ, θi ≥ α2hθ, θi ≥
ω 2||θ||
2.
From here and (3.10), one obtains
M2 ≥ ω 4(m0+ 1)2||h 2||2− ωm2 0 2(m0+ 1)2||h 1||2. (3.11) Estimates for ˜M1. We set h1 = γ1ψ0(¯x) + γ2r′(¯x) + θ1, r(¯x) = νψ0(¯x) + ψ, (3.12) where hθ1, ψ0i = hθ1, r′i = hψ, ψ0i = hψ0, r′i = hψ, r′i = 0 (3.13)
and γ1, γ2 and ν are some constants. As the same line as for M1, we obtain γ12||ψ0||2 = 1 ν2||ψ 0||2 1 2||h|| 2+ Z T 0 ψθ1dx 2 . (3.14) and ˜ M1 ≥ λ2+ λ0(1 + 1d) ||ψ|| 2 ν2 ||ψ0||2 ||θ1||2+ λ 0(1+d) 4ν2 ||ψ0||2||h|| 4 ≥ C1λ2||θ1||2− D1w||h||4. (3.15)
(see the estimates in point C2 of the Appendix2).
We denote ϑ = h1− γ2r′(¯x) = γ1ψ0(¯x) + θ1. Then from (3.14), (3.15) and (since
λ2 = −ω + √ 4ω2+ 6c and λ 2 = −ω − √ 4ω2+ 6c the inequalities λ 2 ≤ 13|λ0| ≤ w, we have ||ϑ||2 = γ2 1||ψ0||2+ ||θ1||2 ≤ 1 + (d+1)||ψ||dν2 2 ||ψ0||2 ||θ1||2+ 4ν21+d ||ψ0||2||h|| 4 ≤ C2||θ1||2+ D2ω− 1 2||h||4. Then ||θ1||2 ≥ ||ϑ|| 2 C2 − D2||h||4 C2ω 1 2 and hence, by (3.15) and λ2 ≤ w,
M1 ≥ C1λ2 C2 ||ϑ|| 2 −C1D2C+ C2D1 2 ω12||h||4 = C3λ2||ϑ||2− D3ω 1 2||h||4. (3.16)
After differentiating (2.8) with respect to ξ and using (3.3), we obtain 0 = 2ReR0T eiη(ϕxxhx+ qϕxh)dx = 2R0T[(−ω + 3v2+ 3r2+ q)r′h 1+ v(−3ω + v2+ 3r2+ q)rh2]dx. From (3.9), we have Z T 0 qrh2dx = − Z T 0 [2vr′h1 + (−ω + v2+ r2)rh2]dx
and replacing in the above equality, we obtain Z T
0 [(−ω + v
2+ 3r2+ q)r′h
1+ v(−2ω + 2r2)rh2]dx = 0.
Substituting h1 = γ2r′(¯x) + ϑ in the above equality and using the orthogonality
condition hr′, ϑi = hr′, γ 1ψ0+ θ1i = 0, we obtain γ2||r′||2 −ω + v2+ q + 3||rr′|| 2 ||r′||2 + Z T 0 [2v(−ω + r 2)rh 2+ 3r2r′ϑ]dx = 0.
Using that 3||rr||r′||′||22 ≥ ω (see the estimates in point B2 of the Appendix2), we further have |γ2| ||r′|| ≤ R0T [2v(−ω + r2)rh2+ 3r2r′ϑ]dx (v2+ q)||r′|| ≤ 2|v| || − ωr + r 3|| · ||h 2|| + 3||r2r′|| · ||ϑ|| (v2+ q)||r′|| ≤ m0(||ϑ|| + ||h2||). Hence ||h1|| ≤ |γ2| ||r′|| + ||ϑ|| ≤ (m0+ 1)||ϑ|| + m0||h2||, which yields ||ϑ||2≥ ||h1|| 2 2(m0+ 1)2 − m0 m0+ 1 2 ||h2||2.
Replacing in (3.16), we finally obtain
M1 ≥ C3λ2 2(m0+ 1)2||h1|| 2 − C3λ2m 2 0 (m0+ 1)2||h2|| 2 − D3ω 1 2|h||4. (3.17)
The estimate for ∆M.
From (3.11) and (3.17), one obtains
M1+ M2 ≥ C3λ2− ωm20 2(m0+ 1)2 ||h1|| 2+ω − 4C3λ2m20 4(m0+ 1)2 ||h2|| 2 − D3ω 1 2||h||4.
We now fix q so that ωm2
0 ≤ 12C3λ2 and assuming that C3 ≤ 1
2 (which is no loss of
generality), one has also 4C3λ2m20 ≤ 12ω. Therefore we come to
M1 + M2 ≥ C4λ2(||h1||2+ ||h2||2) − D3ω 1 2||h||4 = C 4λ2||h||2− D3ω 1 2||h||4 where C4 and D3 are absolute positive constants independent on the parameters of
the system.
On the other hand, estimating directly I2 from below (for this purpose we use its
initial formula), we have I2 ≥ ||hx||2+ RT 0 (ω + v 2− 2r2)|h|2dx − 2|v|RT 0 |h| · |hx|dx − RT 0 r 2|h|2dx ≥ ||hx||2+ (ω + v2− 2r02)||h||2− 2v2||h||2−12||hx|| 2− r2 0||h||2 = 12||hx||2− (v2+ 5ω)||h||2.
Similarly, |I3| ≤ max(4|2ω| 1 2|h| + |h|2)||h||2. Let 0 < m < 1 2. We obtain ∆M = 2mI2+ (1 − 2m)(M1+ M2) + I3 ≥ m||hx||2− 2m(v2+ 5ω)||h||2+ (1 − 2m)(C4λ2||h||2− D3ω 1 2||h||4) − max(4|2ω|12|h| + |h|2)||h||2 = m||hx||2+ [−2m(v2+ 5ω) + (1 − 2m)C4λ2] ||h||2 −[max(4|2ω|12|h| + |h|2) + (1 − 2m)D 3ω 1 2||h||2]||h||2. We choose m, so that 2qm = (1 − 2m)C4λ2− 2m(v2+ 5ω), i.e. 2m = C4λ2 q + v2+ 5ω + C 4λ2 < 1. From the inequality
|h|2 ≤ T1 Z T 0 |h| 2dx + 2 Z T 0 |h| 2dx Z T 0 |h x|2dx 1 2 we obtain |h|2 ≤ T1 Z T 0 |h| 2dx +√q Z T 0 |h| 2dx + 1 √q Z T 0 |h x|2dx.
Hence for sufficiently large q, we obtain
max |h(x, t)|2 ≤ √2qd2q(u, ϕ)
and moreover ||h||2 ≤ q−1d2
q(u, ϕ). Consequently we can choose δ0 > 0, such that for
dq(u, ϕ) < δ0, we will have [max(4|2ω|
1
2|h| + |h|2) + (1 − 2m)D
3ω
1
2]||h||2 ≤ qm. Finally, we obtain that if dq(u, ϕ) < δ0, then ∆M ≥ md2q(u, ϕ). Proposition 3.1 is
completely proved. ✷
Proof of Theorem 3.1. The proof of the theorem in the case P (u) = ||u||2 = ||ϕ||2 = P (ϕ) is the same as in Theorem 2.1. If P (u) 6= P (ϕ), we proceed similarly
as in Theorem 2.1. We have ||ϕ|| = (2√2r0E(k))1/2, where r0 is given by (3.5).
We claim there are respective parameter values ω∗, c∗, and corresponding ϕ∗, r∗, r∗ 0,
k∗, see (3.5) and (3.6), such that ϕ∗ has a period T in x and moreover, 2√2r∗
0E(k∗) =
||u||2. By (3.5), we obtain the equations
2√2K(k∗)
r∗
0 − T = 0,
2√2r∗0E(k∗) − ||u||2 = 0.
(3.18)
If (3.18) has a solution k∗ = k∗(T, ||u||), r∗
0 = r0∗(T, ||u||), then the parameter values
we need are given by
Moreover, one has ||ϕ∗|| = ||u|| and we could use the restricted result we established
above. As k = k∗(T, ||ϕ||), r
0 = r∗0(T, ||ϕ||), it remains to apply the implicit function
theorem to (3.18). Since the corresponding Jacobian determinant reads 2√2K′(k∗) r∗ 0 − 2√2K(k∗) r∗ 0 2 2√2r∗ 0E′(k∗) 2 √ 2E(k∗) = 8 r∗ 0 (KE)′ > 0
the existence of ω∗ and c∗ with the needed properties is established.
By (3.18) and our assumption, we have 2√2K(k∗) r∗ 0 = 2 √ 2K(k) r0 = T. (3.19)
Choosing η = 2(ω∗− ω)t, ξ = 0, we obtain the inequality
d2(ϕ∗, ϕ) ≤ (1 + v2)||r∗− r||2+ ||r∗′ − r′||2
For Φ(ρ) = ρdn(zρ, k(ρ)), where k = k(ρ) is determined from K(k) = 12ρT , we get r∗ − r = Φ(r∗
0) − Φ(r0) = (r0∗ − r0)Φ′(ρ) with some appropriate ρ. Moreover,
|Φ′(ρ)| ≤ C 0. Hence |r∗− r| ≤ C0|r0∗− r0|, |r∗ ′ − r′| ≤ C 1|r∗0− r0|, and by (3.19) d(ϕ∗, ϕ) ≤ |r0∗− r0| ≤ 2√2C T |K(k ∗) − K(k)| ≤ 2 √ 2C T |K ′(k)||k∗− k|.
From the inequalities
|||ϕ∗|| − ||ϕ||| = |||u|| − ||ϕ||| ≤ d(u, ϕ)|t=0< δ it follows that −2√2r0E −1/2 δ < (||ϕ||)−1||ϕ∗|| − 1 < 2√2r0E −1/2 δ and, therefore, 1 − δ1 < r ∗ 0E(κ∗)| r0E < 1 + δ1, i.e. |r ∗
0E(κ∗) − r0E| < r0Eδ1, where
δ1 = (1 + (2r0E)−1/2δ)2− 1. On the other hand
|r∗ 0E(k∗)−r0E(k)| = 2√2 T |K(k ∗)E(k∗)−K(k)E(k)| = 2 √ 2 T |(KE) ′(k)||k∗−k| ≥ C 2|k∗−k|,
with appropriate C2 > 0. Thus
d(u, ϕ∗)|t=0≤ d(u, ϕ)|t=0+ d(ϕ, ϕ∗)|t=0< δ + r0E eCδ1 = δ0.
4
Appendix 1
Below, we provide some estimates needed in our proofs concerning the quadratic Schr¨odinger equation. Without loss of generality, we will assume that r > 0. The case of negative r is dealt with by changing r → −r in all equations from Section 1.
For n ∈ Z and c ∈ (−13ω3, 0), consider the line integrals In(c) and their derivatives
I′ n(c) given by In(c) = I H=c xnydx, In′(c) = I H=c xndx 2y (4.1)
where the integration is along the right oval contained in the level set {H = c} and H(x, y) = y2− ωx2+2
3x
3. These integrals would be useful because
Z T 0 rn(t)dt = 2 Z 1 2T 0 rn(t)dt = 2 Z r0 r1 xndx q c + ωx2 −2 3x3 = I H=c xndx y = 2I ′ n(c). (4.2)
(we applied a change of the variable r(t) = x in the integral and used equation (2.3)). The properties of In are well known, see e.g. [9] for a similar treatment. Below, we
list some facts we are going to use.
Lemma. (i) The following identity holds:
3ncIn−1+ 3ω(n + 3)In+1− (2n + 9)In+2= 0, n ∈ Z
which implies
I2 = ωI1, I3 = 113cI0 +1211ω2I1, I4 = 14345cωI0+ (136c + 180143ω3)I1, (4.3)
(ii) The integrals I0 and I1 satisfy the system
6cI′
0+ 2ω2I1′ = 5I0,
6ωcI′
0+ (30c + 12ω3)I1′ = 35I1.
(iii) The ratio R(c) = I′
1(c)/I0′(c) satisfies the Riccati equation and related system
6c(3c + ω3)R′(c) = ω2R2(c) + 6cR(c) − 3ωc, ˙c = 6c(3c + ω
3),
˙
R = ω2R2+ 6cR − 3ωc, (4.4)
which imply estimates
− 3c ω2 ≤ R(c) ≤ 5ω 6 − c 2ω2. (4.5)
The equations in (i)–(iii) are derived in a standard way, see [9] for more details. The estimates (4.5) follow from the fact that, in the (c, R)-plane, the graph of R(c) coincides with the concave separatrix trajectory of the system (4.4) contained in the
triangle with vertices (0, 0), (−1 3ω
3, ω) and (0,5ω
6 ) and connecting the first two of
them.
After this preparation, we turn to prove the estimates we used in the preceding Section 2.
A1. The estimate for A1 = ||r
3/2||2
||r||2 . By (4.2), (4.3) and the first inequality in (4.5), we have A1 = RT 0 r 3dt RT 0 r2dt = I3′ I′ 2 = 3 5 cI′ 0+ 2ω2I1′ ωI′ 1 = 3c 5ωR + 6ω 5 ≥ ω.
B1. The estimate forB1 = 2||r
1/2
r′||2
||r′||2 . By (2.3) and (4.3), we have as above
B1 = 2R0T r(c + ωr2−2 3r 3)dt RT 0 (c + ωr2− 2 3r3)dt = 2(cI ′ 1+ ωI3′ −23I4′) cI′ 0+ ωI2′ − 23I3′ = 6 7 cωI′ 0+ (5c + 2ω3)I1′ 3cI′ 0+ ω2I1′ = 6 7 cω + (5c + 2ω3)R 3c + ω2R ≥ 12 7 ω. To obtain the last inequality, we used both estimates in (4.5).
C1. The estimate for C1 = λ2+ λ0(1 + 1d) ||ψ||
2 ν2 ||ψ0||2. By (2.14) and (2.15) we have ||ψ||2 = ||r||2− ν2||ψ0||2, where ν = hr, ψ0i ||ψ0||2 . Therefore C1 = λ2+ λ0 1 + 1 d ||r||2||ψ 0||2 hr, ψ0i2 − 1 .
Below, we need to use the following well-known equalities (see e.g. [7]) which are written out here for reader’s convenience:
(i) R0Kdn2xdx = E(k) (ii) R0Kdn4xdx = 4−2k2 3 E(k) + k2 −1 3 K(k) (iii) R0Kdn6xdx = 8k4 −23k2 +23 15 E(k) + 4(k2 −1)(2−k2 ) 15 K(k) (iv) R0Kdn2xsn2xdx = 2k2 −1 3k2 E(k) + 1−k 2 3k2 K(k) (iv) R0Kdn2xsn4xdx = 8k4 −3k2 −2 15k4 E(k) + 2(1−k2 )(1+2k2 ) 15k4 K(k) (v) R02Kcn2xdx = 2 k2[E(k) + (k2− 1)K(k)] (vi) R02Ksn2xdx = 2 k2[K(k) − E(k)]
(vii) R0Kdnxdx = R02Kdnxsn2xdx =R2K 0 dnxcn 2xdx = π/2 (viii) R02Kdnxsn2xcn2xdx = π/8 (ix) R02πdn3xdx = 2−k2 2 π
Now, we first recall that
r(x) = α2[1 + k2+√1 − k2+ k4− 3k2sn2(αx; k)],
ψ0(x) = dn(x; k)[1 − (1 + 2k2−
√
1 − k2+ 4k4)sn2(x; k)] > 0.
From (2.2), we obtain after integration
||r||2= w Z T 0 r(x)dx = wr1T + w(r0− r1) Z T 0 cn2(αx)dx. Then, by using (v) and the expressions of ω, r1 etc, we calculate
||r||2 = 16α3√1 − k2+ k4[(k2
− 2 +√1 − k2+ k4)K(k) + 3E(k)]
Similarly, by direct calculations and making use of (i)-(ix), we come to the expressions
||ψ0||2 = 4 α √ 1 − k2+ 4k4 15k4 {(k 2− 1)[(8k4+ 3k2+ 2) − 2(2k2+ 1)√1 − k2+ 4k4]K(k) +[(8k4− 3k2− 2)√1 − k2+ 4k4− 2(8k6 − 4k4− k2− 1)]E(k)}. hr, ψ0i = πα[(1 + k2 + √ 1 − k2 + k4)(1 − 2k2+√1 − k2+ 4k4) −9k 2 4 ( 1 3 − 2k 2+√ 1 − k2+ 4k4)]
Next, calculating the asymptotical expansions near k = 0, we obtain ||r||2 = 16πα3(1 − 3 4k 2 +39 64k 4+ . . .) hr, ψ0i = 4πα(1 − 74k2+7132k4+ . . .) ||ψ0||2 = π α √ 1 − k2+ 4k4(1 − 9 4k 2+135 64k 4+ . . .) −λ0 λ2− λ0 = 1 − 1 2k 2+15 16k 4+ . . . √ 1 − k2+ 4k4 and, finally, −λ0 λ2− λ0 ||r||2||ψ 0||2 hr, ψ0i2 = 1 − 21 32k 4 + O(k6). (4.6) Let us denote by 1−δ the right-hand side of (4.6). Clearly, δ = δ(k) satisfies 1 > δ > 0 for k small enough, say 0 < k ≤ k0. Fixing such a small k, then (4.6) yields
||r||2||ψ 0||2 hr, ψ0i2 = (1 − δ) 1 − λλ2 0 ≤ 1 + λλ2 0(δ − 1).
Therefore C1 = λ2+ λ0 1 + 1 d ||r||2||ψ 0||2 hr, ψ0i2 − 1 ≥ λ2 1 + 1 + 1 d (δ − 1) = Cλ2
where C is a positive constant, provided that d is chosen sufficiently large. Note that the above estimate is not uniform in k when k tends to zero. This is because δ = O(k4) and hence d = O(k−4).
When k ≥ k0, we can simply draw the graph of the corresponding function f (k) = ||r||2
||ψ0||2
hr,ψ0i2 −1+
λ2
λ0 to see that it is negative and placed far from zero. Hence, C1 ≥ Cλ2, too (uniformly for k ≥ k0).
0.2 0.4 0.6 0.8 1 -0.5 -0.4 -0.3 -0.2 -0.1 Figure 2: Graph of f (k) < 0
D1. The estimates for D11 = ||r
2
−ωr||
||r′|| and D12 =
||r′||
||r||. Making use of statements
(i) and (ii) of the Lemma, we have
D112 = RT 0 (r 2− ωr)2dt RT 0 (c + ωr2− 2 3r3)dt = ω 2I′ 2− 2ωI3′ + I4′ cI′ 0+ ωI2′ − 23I3′ = 5 7. −3cωI′ 0+ (6c + ω3)I1′ 3cI′ 0+ ω2I1′ = −ωI0+ 2I1 I0 ≤ 2ω.
The last inequality follows from the fact that the periodic trajectories of the Hamil-tonian system lie inside the saddle loop, in the domain H(x, y) < 0 which implies y2 < x2(ω − 2 3x) and x < 3 2ω. Similarly, D122 = RT 0 (c + ωr2− 2 3r3)dt RT 0 r2dt = cI ′ 0+ ωI2′ − 23I3′ I′ 2 = 3cI0′ + ω 2I′ 1 5ωI′ 1 = (3c + ω 3)I 0 ω(7I1− ωI0) ≤ 3c + ω3 5ω2 .
The last inequality is due to the known fact [14] (Theorem 12), that the function Q(c) = I1(c)/I0(c) is strictly decreasing, therefore Q(c) > Q(0) = 67ω, the value
obtained by direct calculation of elementary integrals.
5
Appendix 2
First of all, let us mention that the operator L1, defined by differential expression
(3.8) and equipped with periodic boundary conditions in [0, T ], has the following spectral data λ0 = −ω − √ 4ω2+ 6c, ψ 0 = 3r2− 3ω − λ0, λ1 = 0, ψ1 = r′, λ2 = −ω + √ 4ω2+ 6c, ψ 2 = 3r2− 3ω − λ2. (5.1) For n ∈ Z and c ∈ (−1 2ω
2, 0), consider the line integrals I
n(c) and their derivatives
I′ n(c) given by In(c) = I H=c xnydx, In(c) = I H=c xndx 2y (5.2)
where one can assume for definiteness that the integration is along the right oval contained in the level set {H = c}, H(x, y) = y2 − ωx2 + 12x4. As above, these integrals satisfy I′
n(c) = 12
RT 0 r
n(z)dz. The properties of I
n are well known. We only
list some facts we are going to use.
Lemma. (i) The following identity holds:
(n + 6)In+3− 2ω(n + 3)In+1− 2ncIn−1= 0, n ∈ Z
which implies I3 = ωI1, I4 = 2c 7I0+ 8ω 7 I2, I6 = 8ωc 21 I0+ 2c 3 + 32ω2 21 I2. (5.3)
(ii) The integrals I0 and I2 satisfy the system
4cI′
0+ 2ωI2′ = 3I0,
4ωcI′
0+ (12c + 8ω2)I2′ = 15I2.
(iii) The ratio R(c) = I′
2(c)/I0′(c) satisfies the Riccati equation and related system
(8c2+ 4ω2c)R′(c) = −2ωc + 4cR(c) + ωR2(c), ˙c = 8c
2 + 4ω2c,
˙
R = −2ωc + 4cR + ωR2, (5.4)
which imply estimates
− 2cω ≤ R(c) ≤ −2ωc +3ω
The equations in (i)–(iii) are derived in a standard way, see [9] for more details. The estimates (5.5) follow from the fact that, in the (c, R)-plane, the graph of R(c) coincides with the concave separatrix trajectory of the system (5.4) contained in the triangle with vertices (0, 0), (−12ω
2, ω) and (0,3
4ω) and connecting the first two of
them. Note that R′(c) = −1/(2ω) at c = −1
2ω2. We also use analyticity of the ratio
R(c) at this point and properties of the phase portrait of (5.4) to verify the above statements.
After this preparation, we turn to prove the estimates we used in the preceding sections.
A2. The estimate for A2 = ||r
2
||2
||r||2 . By (4.2), (5.3) and the first inequality in (5.5), we have A2 = RT 0 r4dt RT 0 r2dt = I4′ I′ 2 = 2cI0′ + 4ωI2′ 3I′ 2 = 2c 3R + 4ω 3 ≥ ω.
B2. The estimate forB2 = ||rr
′||2 ||r′||2 . As I6′ = 16ωc 15 I ′ 0+ 6c 5 + 32ω2 15 I2′, we have by (3.4) as above B2 = RT 0 r2(c + ωr2− 1 2r4)dt RT 0 (c + ωr2− 1 2r4)dt = 2cI2′ + 2ωI4′ − I6′ 2cI′ 0+ 2ωI2′ − I4′ = 2 5 ωcI′ 0+ (3c + 3ω2)I2′ 2cI′ 0+ ωI2′ = 2 5 ωc + (3c + 2ω2)R 2c + ωR ≥ 4ω 5 . To obtain the last inequality, we used both estimates in (5.5).
C2. The estimate for C2 = λ2+ λ0(1 + 1d) ||ψ||
2 ν2 ||ψ0||2. By (3.12) and (3.13) we have C2 = λ2+ λ0 1 + 1 d ||r||2||ψ 0||2 hr, ψ0i2 − 1 . Next, hr, ψ0i = Z T 0 [3r3− (3ω + λ0)r]dt = 6I3′ − (6ω + 2λ0)I1′ = −2λ0I1′, ||r||2||ψ 0||2 = RT 0 r2dt RT 0 (3r2− 3ω − λ0)2dt = 4I′ 2[9I4′ − (18ω + 6λ0)I2′ + (3ω + λ0)2I0′] = 4I′ 2[(6c + (3ω + λ0)2)I0′ − (6ω + 6λ0)I2′]. By (3.5), we have I2′(c) = 1 2 Z T 0 r2dt = (r02/α) Z K(k) 0 dn2(t)dt =√2r0E(k). (5.6)
Making use of the identity E(k) = 1 2πF ( 1 2, − 1 2, 1, k
2) where F is the Gauss
hyperge-ometric function, we obtain an appropriate expansion to estimate E from above E(k) = π 2 1 − k 2 4 − 3k4 64 − 5k6 512 − . . . , , E2(k) ≤ π 2 4 1 − k 2 2 − k4 32
with all removed terms negative. As I′
1 = √12π, by (3.5) this implies I2′2≤ −I1′2ω 2− 10ωr2 0+ r04 8r2 0 . Together with I′ 0I2′ ≥ I1′2, this yields ||r||2||ψ 0||2 hr, ψ0i2 − 1 ≤ 1 λ2 0 6c + (3ω + λ0)2+ 3 4r2 0 (ω + λ0)(ω2− 10ωr02+ r04) − 1 = λ2 λ0 λ2+ ω 4r2 0 − 1 ≤ λλ2 0 √ 3 8 − 1 !
where the equality is obtained by direct calculations. Therefore,
C2 ≥ λ2 − 1 d + d + 1 d √ 3 8 ! = ¯C2λ2
with ¯C2 > 0 an absolute constant when d ≥ 4 is fixed.
As a by-product of our calculations, we easily obtain also the estimate λ0 ν2||ψ 0||2 = λ0||ψ0|| 2 hr, ψ0i2 ≥ λ2( √ 3 8 − 1) + λ0 ||r||2 ≥ −D1ω 1 2.
D2. The estimates for D21 = ||r
3
−ωr||
||r′|| and D22=
||r′||
||r||. We proceed as in case D1
above. Making use of statements (i) and (ii) of the Lemma, we have
D212 = RT 0 (ω 2r2− 2ωr4+ r6)dt RT 0 (c + ωr2− 1 2r4)dt = ω 2I′ 2− 2ωI4′ + I6′ cI′ 0+ ωI2′ − 12I4′ = −4cωI0′ + (18c + 7ω 2)I′ 2 5(2cI′ 0+ ωI2′) = −ωI0+ 3I2 I0 ≤ 5ω. Similarly, D222 = RT 0 (c + ωr 2− 1 2r 4)dt RT 0 r2dt = cI ′ 0+ ωI2′ − 12I4′ I′ 2 = 2cI0′ + ωI2′ 3I′ 2 = (2c + ω 2)I 0 5I2− ωI0 ≤ 2c + ω2 3ω .
As before, we used that I2(c)/I0(c) is a decreasing function and calculated the value
I2(0)/I0(0) = 45ω.
Acknowledgment. The first author has been partially supported by a Research grant DDVU 02/91 (2010) of the Bulgarian Ministry of Education and Science.
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